Difference between revisions of "1983 AIME Problems/Problem 14"
(→Solution 4) |
(→Solution 2) |
||
Line 15: | Line 15: | ||
Doing routine algebra on the above equation, we find that <math>x^2=\frac{65}{2}</math>, so <math>PQ^2 = 4x^2 = \boxed{130}.</math> | Doing routine algebra on the above equation, we find that <math>x^2=\frac{65}{2}</math>, so <math>PQ^2 = 4x^2 = \boxed{130}.</math> | ||
=== Solution 2 === | === Solution 2 === | ||
− | This is a classic [[side chase]] - just set up equations involving key lengths in the diagram. Let the | + | This is a classic [[side chase]] - just set up equations involving key lengths in the diagram. Let the midpoint of <math>QP</math> be <math>M_1</math>, and the midpoint of <math>PR</math> be <math>M_2</math>. Let <math>x</math> be the length of <math>AM_1</math>, and <math>y</math> that of <math>BM_2</math>. |
{{incomplete|solution}} | {{incomplete|solution}} |
Revision as of 23:06, 10 April 2009
Problem
In the adjoining figure, two circles with radii and
are drawn with their centers
units apart. At
, one of the points of intersection, a line is drawn in sich a way that the chords
and
have equal length. (
is the midpoint of
) Find the square of the length of
.
Contents
Solution
Solution 1
First, notice that if we reflect over
we get
. Since we know that
is on circle
and
is on circle
, we can reflect circle
over
to get another circle (centered at a new point
with radius
) that intersects circle
at
. The rest is just finding lengths:
Since is the midpoint of segment
,
is a median of triangle
. Because we know that
,
, and
, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get
. So now we have a kite
with
,
, and
, and all we need is the length of the other diagonal
. The easiest way it can be found is with the Pythagorean Theorem. Let
be the length of
. Then
![$\sqrt{36-x^2} + \sqrt{64-x^2} = \sqrt{56}.$](http://latex.artofproblemsolving.com/6/d/2/6d2150ebb9706e5687fb702d57b9724260acea68.png)
Doing routine algebra on the above equation, we find that , so
Solution 2
This is a classic side chase - just set up equations involving key lengths in the diagram. Let the midpoint of be
, and the midpoint of
be
. Let
be the length of
, and
that of
.
Solution 3
Let . Angles
,
, and
must add up to
. By the Law of Cosines,
. Also, angles
and
equal
and
. So we have
![$\cos^{-1}(x/16)+\cos^{-1}(-11/24)=180-\cos^{-1}(x/12).$](http://latex.artofproblemsolving.com/4/c/6/4c6cea3db0103d0c0397ce9207b3afc1f0cd795e.png)
Taking the of both sides and simplifying using the cosine addition identity gives
.
Solution 4
Let the circles of radius and
be centered at
and
respectively. Let the midpoints of
and
be
and
Dropping a perpendicular from
to
(let the point be
) gives a rectangle.
Now note that triangle is right. Let the midpoint of
(segment of length
) be
Hence,
By now obvious similar triangles, so it's a quick system of two linear equations to solve for the desired length.
See also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |